Question

if the sum of first P terms of an ap is equal to the sum of first Q terms then show that the sum of its first term P + Q terms is zero

Answer 1

Sum of first p terms of an AP is�

Sum of first q terms of an AP is�

Answer 2

    We know that for an A.P, Sn = n2[2a+(n−1)d]n2[2a+(n−1)d]

∴                                             p2[2a+(p–1)d]p2[2a+(p–1)d]  = q2[2a+(q−1)d]q2[2a+(q−1)d]

          Multiplying by 2 on both the sides, we get;
          p [(2a + (p – 1)d)] = q [(2a + (q – 1)d]
∴       2ap + p(p – 1)d = 2aq + q(q – 1)d
∴       2ap – 2aq + p(p – 1)d – q(q – 1)d = 0
∴       2a (p – q) + d[p(p – 1) – q(q – 1)] = 0
∴       2a(p – q) + d(p2 – p – q2 + q) = 0
∴       2a(p – q) + d(p2 – q2 – p + q) = 0
∴       2a(p – q) + d[(p + q) (p – q) – (p – q)] = 0
∴       2a(p – q) + d(p + q) (p – q) – d(p – q) = 0
         p ≠ q                                                                                                     [Given]
∴      p – q ≠ 0
        Dividing throughout by (p – q), we get;
                                    2a + d(p + q) – d  =  0                                                ...(I)
                                     2a + d(p + q – 1) = 0
        We have to prove S(p + q) = 0

∴       S(p + q) = p+q2[2a+(p+q−1)d]p+q2[2a+(p+q−1)d] 

∴       S(p + q) = p+q2∗(0)p+q2∗(0)                                                                             [From (I)]

∴       S(p + q) = 0

∴     The sum of the first p + q terms is zero.